University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.10 - Related Rates - Exercises - Page 187: 16


${\frac{dP}{dt}}={2RI{\frac{dI}{dt}}+I^2{\frac{dR}{dt}}}$ ${{\frac{dR}{dt}}=-\frac{2P}{I^3}{\frac{dI}{dt}}}=-\frac{2R}{I}{\frac{dI}{dt}}$

Work Step by Step

We know that $P=I^2R$ on differentiating each side with respect to time: ${\frac{dP}{dt}={R{\frac{dI^2}{dt}}+I^2{\frac{dR}{dt}}}}$ ${\frac{dP}{dt}}={2RI{\frac{dI}{dt}}+I^2{\frac{dR}{dt}}}$ when P is constant, its derivative is zero, so the above equation becomes: $0={2RI{\frac{dI}{dt}}+I^2{\frac{dR}{dt}}}$ ${2RI{\frac{dI}{dt}}=-I^2{\frac{dR}{dt}}}$ ${{\frac{dR}{dt}}=-\frac{2R}{I}{\frac{dI}{dt}}}$ ${{\frac{dR}{dt}}=-\frac{2RI}{I^2}{\frac{dI}{dt}}}$ since $\frac{P}{I}= IR$ ${{\frac{dR}{dt}}=-\frac{2P}{I^3}{\frac{dI}{dt}}}$ The final answer: ${\frac{dP}{dt}}={2RI{\frac{dI}{dt}}+I^2{\frac{dR}{dt}}}$ ${{\frac{dR}{dt}}=-\frac{2P}{I^3}{\frac{dI}{dt}}}=-\frac{2R}{I}{\frac{dI}{dt}}$
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